Let [math]P[/math] be a point on the hyperbola [math]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/math] with foci [math]F_1[/math] and [math]F_2[/math].[br]It must satisfy [math]PF_1-PF_2=2a[/math].[br][br]Choose point [math]G[/math] on [math]\overline{F_1P}[/math] such that [math]GP=PF_2[/math].[br]Note that [math]F_1G=PF_1-GP=PF_1-PF_2=2a[/math].[br]For midpoint [math]M[/math] of  [math]\overline{GF_2}[/math], the line [math]\overline{PM}[/math] is perpendicular to [math]\overline{GF_2}[/math][br]and all four of the marked angles around [math]P[/math] are equal.[br][math][/math][br]Let [math]Q[/math] be any point on [math]\overline{PM}[/math] other than [math]P[/math] and connect it to [math]F_1[/math], [math]F_2[/math] and [math]G[/math].[br][math]QF_2=QG[/math] because [math]QM[/math] is a perpendicular bisector of [math]\overline{GF_2}[/math].[br][math]F_1G+QF_2=F_1G+QG>QF_1[/math] by the triangle inequality.[br][br]Since [math]2a=F_1G>QF_1-QF_2[/math], [math]Q[/math] must lie outside the hyperbola,[br]so [math]\overline{PM}[/math] never passes through the hyperbola and must be tangent to it.[br][br]It follows that a ray [b]from[/b] focus [math]F_1[/math] will reflect off the hyperbola directly [b]away[/b] from focus [math]F_2[/math].[br]Similarly, a ray directed [b]to[/b] focus [math]F_2[/math] will reflect off the hyperbola [b]toward[/b] focus [math]F_1[/math].